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22 Mar 2026, 19:00
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E
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Difficulty:
75%
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Question Stats:
51% (02:13) correct
49%
(02:20)
wrong
based on 47
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During an experiment, the population of a bacterium reduced by a factor of ‘k’ every minute. The population of the bacterium ten minutes after the beginning of the experiment was 80,000. Another 5 minutes later, the population was 64,000. What was the population of the bacterium at the beginning of the experiment?
A. 96,000
B. 100,000
C. 112,000
D. 120,000
E. 125,000
A. 96,000
B. 100,000
C. 112,000
D. 120,000
E. 125,000
22 Mar 2026, 22:45
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We can see 10 mins post beginning the experiment amount of bacteria = 80000
After another 5 mins it reduced to 64,000
If we see this from below, the rate at which the increase has happened = 25% [ie., (80000-64000)/64000]
We can say that every 5 mins the amount of bacteria increases by 25%
Therefore, 5 min after the start of the experiment the amount of bacteria = 80000 * 1.25 = 100000
Therefore, the amount of bacteria present at the start of the experiment = 100000 * 1.25 = 125000
Option E
After another 5 mins it reduced to 64,000
If we see this from below, the rate at which the increase has happened = 25% [ie., (80000-64000)/64000]
We can say that every 5 mins the amount of bacteria increases by 25%
Therefore, 5 min after the start of the experiment the amount of bacteria = 80000 * 1.25 = 100000
Therefore, the amount of bacteria present at the start of the experiment = 100000 * 1.25 = 125000
Option E
22 Mar 2026, 23:25
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Let N be population at beginning, it reduces by factor of K i.e. after 1 min N(1-1/k)
So population at end of t min will be N(1 - 1/k) ˆt
Given, N(1-1/k)ˆ10 = 80000 and N(1-1/k)ˆ15=64000
To solve, Cube the 1st exp and square the 2nd exp, divide both, N = (80)ˆ3 x 1000/(64)ˆ2 = 125000
E
So population at end of t min will be N(1 - 1/k) ˆt
Given, N(1-1/k)ˆ10 = 80000 and N(1-1/k)ˆ15=64000
To solve, Cube the 1st exp and square the 2nd exp, divide both, N = (80)ˆ3 x 1000/(64)ˆ2 = 125000
E
